Answer:
Part a)
v = 16.52 m/s
Part b)
v = 7.47 m/s
Explanation:
Part a)
(a) when the large-mass object is the one moving initially
So here we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]
since this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]
[tex]v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}[/tex]
[tex]v = 16.52 m/s[/tex]
Part b)
(b) when the small-mass object is the one moving initially
here also we can use momentum conservation as the net force on the system of two masses will be zero
so here we can say
[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]
Again this is a perfect inelastic collision so after collision both balls will move together with same speed
so here we can say
[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]
[tex]v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}[/tex]
[tex]v = 7.47 m/s[/tex]
A golfer hits a shot to a green. The ball leaves the club at a speed of 20 m/s at an angle 32° above the horizontal. It rises to its maximum height and then falls down to the green. What is the speed of the ball at maximum height? Ignore air resistance.
Answer:
16.96 m/s
Explanation:
It is the case of projectile motion in which the projectile fires from the ground and again it hit the ground.
The angle of projection is 32 degree and the velocity of projection is 20 m/s
he velocity at the maximum height is equal to the horizontal component of velocity which always remains constant as there is no acceleration along x axis.
Velocity at the highest point = u Cos 32 = 20 x Cos 32 = 16.96 m/s
An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to the local atmospheric pressure) in kPa inside of the diving chamber required to keep water from entering the chamber. (Assume SG=1.03)
Answer:
Gauge Pressure required = 606.258 kPa
Explanation:
Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.
Thus at a depth of 60m we have pressure of water equals
[tex]P(z)=P_{0}+\rho _wgh[/tex]
Now the gauge pressure is given by
[tex]P(z)-P_{0}=\rho _wgh[/tex]
Applying values we get
[tex]P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa[/tex]
Final answer:
To keep water from entering an open diving chamber at 60 meters depth with sea water of specific gravity 1.03, the air pressure inside the chamber must be equal to the water pressure at that depth, which is 603.84 kPa.
Explanation:
To calculate the air pressure (gage pressure) inside the diving chamber that prevents water from entering, we will use the pressure-depth relationship in fluids. The gage pressure is the pressure relative to atmospheric pressure. Given that the chamber is at a depth of 60 meters underwater and the specific gravity (SG) of the sea water is 1.03, we know that the sea water is denser than pure water (whose SG is 1). To prevent water from entering the chamber, the air pressure inside the chamber must be equal to the water pressure at that depth. The pressure due to a column of liquid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the liquid, g is the acceleration due to gravity (9.8 m/s²), and h is the height (or depth) of the liquid column.
Since the water's specific gravity is 1.03, its density (ρ) can be calculated as ρ = SG * ρwater, where ρwater is the density of pure water (1000 kg/m³). So, ρ = 1.03 * 1000 kg/m³ = 1030 kg/m³. The depth (h) is 60 meters.
Now, we can calculate the pressure:
P = ρgh = (1030 kg/m³)(9.8 m/s²)(60 m) = 603,840 Pa or N/m².
To convert this to kilopascals (kPa), we divide by 1,000:
P = 603.84 kPa
This is the gage pressure inside the chamber required to prevent water from entering.
An automobile tire has a volume of 1.63 x 10-2 m3 and contains air at a gauge pressure (pressure above atmospheric pressure) of 165 kPa when the temperature is 0.00°C. What is the gauge pressure of the air in the tires when its temperature rises to 27.3°C and its volume increases to 1.70 x 10-2 m3
Answer: The gauge pressure of the air in the tires is 179.5 kPa.
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = Atmospheric pressure + gauge pressure = 101 kPa + 165 kPa = 266 kPa
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = [tex]1.63\times 10^{-2}m^3[/tex]
[tex]V_2[/tex] = final volume of gas = [tex]1.70\times 10^{-2}m^3[/tex]
[tex]T_1[/tex] = initial temperature of gas = [tex]0^oC=273+0=273K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]27.3^oC=273+27.3=300.3K[/tex]
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]\frac{266\times 1.63\times 10^{-2}}{273K}=\frac{P_2\times 1.70\times 10^{-2}}{300.3K}[/tex]
[tex]P_2=280.5kPa[/tex]
Gauge pressure = Absolute pressure - atmospheric pressure = (280.5 - 101) kPa= 179.5 kPa
Therefore, the gauge pressure of the air in the tires is 179.5 kPa.
A boy throws a ball straight up in the air with an initial velocity of V+3 mls . The ball reaches a maximum height and starts towards earth. The boy catches the ball at the same height that it left his hand. How long is the ball in the air? 6. a.) =1.01 s b.) t-0.61s c.) 1-2.32s d.) t=1.23h e.) None of the above.
Answer:
Option (b)
Explanation:
Let the ball takes time t to reach the maximum height. The final velocity at maximum height is zero.
Use I equation of motion, we get
V = u + gt
0 = 3 - 9.8 x t
t = 0.306 s
As the air friction is negligible, so the ball takes same time to reach the hand of boy. Thus the time taken by the ball in air is
T = 2 t = 2 x 0.306 = 0.61 s
What is the resistance of a 1000 m long copper wire with a 1 × 10-6 m² (1 mm²) cross-section assuming copper's resistivity is 1.68 × 10-8 Ω·m? Answer in the form: ##.#
Answer:
[tex]R=16.8[/tex]Ω
Explanation:
l=1000m
a=1mm2
p=1.68x10^-8
Electrical Resistivity Equation
[tex]R=p(\frac{L}{A} )\\[/tex]
where
p= proportional constant ρ (the Greek letter “rho”) is known as Resistivity.
L = is the length in metres (m)
A= is the area in square metres (m2),
R=1.68×10^-8×(1000/1×10^-6)
[tex]R=16.8[/tex]Ω
The Eiffel tower in Paris is 300 meters tall on a cold day (T = -24 degrees Celsius), what is its height on a hot day when the temperature is 35 degrees? (It is made of iron which has a coefficient of linear expansion of 0.000012 per degree Celsius.)
Answer:
Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m
Explanation:
Thermal expansion is given by the expression
[tex]\Delta L=L\alpha \Delta T \\ [/tex]
Here length of Eiffel tower, L = 300 m
Coefficient of thermal expansion, α = 0.000012 per degree Celsius
Change in temperature, = 35 - (-24) = 59degrees Celsius
Substituting
[tex]\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\ [/tex]
Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9.0 m/s, they then have the same kinetic energy. What were the original speeds of the two cars? Express your answers using two significant figures separated by a comma?
Answer:
The original speeds of the two cars were :
[tex]6.36\frac{m}{s},12.72\frac{m}{s}[/tex]
Explanation:
Let's start reading the question and making our equations in order to find the speeds.
The first equation is :
[tex]m_{1}=2m_{2}[/tex] (I)
The kinectic energy can be calculated using the following equation :
[tex]K=(\frac{1}{2}).m.v^{2}[/tex] (II)
Where ''K'' is the kinetic energy
Where ''m'' is the mass and where ''v'' is the speed.
By reading the exercise we find that :
[tex]K_{1}=\frac{K_{2}}{2}[/tex] (III)
If we use (II) in (III) :
[tex](\frac{1}{2}).m_{1}.v_{1}^{2}=(\frac{1}{2}).m_{2}.v_{2}^{2}.(\frac{1}{2})[/tex]
[tex]m_{1}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex] (IV)
If we replace (I) in (IV) ⇒
[tex]2.m_{2}.v_{1}^{2}=\frac{m_{2}.v_{2}^{2}}{2}[/tex]
[tex]4.v_{1}^{2}=v_{2}^{2}[/tex]
[tex]2.v_{1}=v_{2}[/tex] (V)
'' When both cars increase their speed by [tex]9.0\frac{m}{s}[/tex], they then have the same kinetic energy ''
The last equation is :
[tex](\frac{1}{2}).m_{1}.(v_{1}+9)^{2}=(\frac{1}{2}).m_{2}.(v_{2}+9)^{2}[/tex] (VI)
If we use (I) in (VI) ⇒
[tex]2.m_{2}.(v_{1}+9)^{2}=m_{2}.(v_{2}+9)^{2}[/tex]
[tex]2.(v_{1}+9)^{2}=(v_{2}+9)^{2}[/tex]
If we use (V) in this last expression ⇒
[tex]2.(v_{1}+9)^{2}=(2.v_{1}+9)^{2}[/tex]
[tex]2.(v_{1}^{2}+18v_{1}+81)=4v_{1}^{2}+36v_{1}+81[/tex]
[tex]2v_{1}^{2}+36v_{1}+162=4v_{1}^{2}+36v_{1}+81[/tex]
[tex]2v_{1}^{2}=81[/tex]
[tex]v_{1}^{2}=40.5[/tex]
[tex]v_{1}=\sqrt{40.5}=6.36[/tex]
We find that the original speed [tex]v_{1}[/tex] is [tex]6.36\frac{m}{s}[/tex]
If we replace this value in the equation (V) ⇒
[tex]2.(6.36\frac{m}{s})=v_{2}[/tex]
[tex]v_{2}=12.72\frac{m}{s}[/tex]
An electromagnetic standing wave has a frequency of 120 MHz. (3396) Problem 2: What is the distance (in m) between adjacent antinodes of this standing wave?
Answer:
The distance between adjacent antinodes of the standing wave is 1.25 m.
Explanation:
Given that,
Frequency f= 120 MHz
We need to calculate the distance between adjacent antinodes of the standing wave
Using formula of distance
[tex]\Delta x=\dfrac{\lambda}{2}[/tex].....(I)
We know that,
[tex]\lambda=\dfrac{c}{f}[/tex]
Put the value of [tex]\lambda[/tex] in to the equation (I)
[tex]\Delta x=\dfrac{c}{2f}[/tex]
Where, c = speed of light
f = frequency
Put the all value into the formula
[tex]\Delta x=\dfrac{3\times10^{8}}{2\times120\times10^{6}}[/tex]
[tex]\Delta x=1.25\ m[/tex]
Hence, The distance between adjacent antinodes of the standing wave is 1.25 m.
How much energy is stored by the electric field between two square plates, 8.5 cm on a side, separated by a 2.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 14 nC .
The energy stored by the electric field between two plates, known as a parallel-plate capacitor, is calculated using the respective formulas for voltage, capacitance and energy in a capacitor, factoring in the charge of the plates, surface area, and separation distance.
Explanation:The energy stored by the electric field between two parallel plates (known as a parallel-plate capacitor) can be calculated using the formula for the energy stored in a capacitor, which is U = 0.5 * C * V^2. Here, V is the voltage across the plates and C is the capacitance of the capacitor.
To calculate V, we use the relation V = E * d, where E is the electric field, which can be found using the formula E = Q/A (Charge per Area), and d is the distance separating the plates. In this case, the charge Q is 14 nC, the area A is (8.5 cm)^2 and the distance d is 2.5 mm.
Finally, to find C, we use the formula C = permittivity * (A/d), where the permittivity of free space is approximately 8.85 × 10^-12 F/m. With these variables, C, V and U can be calculated accordingly.
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The energy stored in the electric field of a parallel-plate capacitor can be calculated using the dimensions of the plates, their separation, and charges. For this particular scenario, the energy stored would be approximately 0.44 μJ.
Explanation:In this scenario, we have a parallel-plate capacitor where each plate is a square with side 8.5 cm and has a charge of 14 nC. The plates are separated by a 2.5mm gap. Using this information, we can calculate the energy stored in the capacitor.
Firstly, the surface area A of the plates can be calculated using the formula A = s², where s is the side of the square. Converting 8.5 cm to meters, we get s = 0.085 m. So, A = (0.085 m)² = 0.007225 m².
Next, we calculate the electric field E between the plates using the formula E = Q/ε₀A, where Q is the charge on one plate and ε₀ is the permittivity of free space (8.85 x 10^-12 F/m). Converting 14 nC to coulombs gives Q = 14 x 10^-9 C. Substituting these values into the formula, we find E ≈ 224.717 kV/m.
The energy U stored in an electric field can be calculated using the formula U = 0.5ε₀EA². Substituting the earlier calculated values, we find U ≈ 0.44 μJ.
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The radius of a right circular cone is increasing at a rate of 1.5 in/s while its height is decreasing at a rate of 2.4 in/s. At what rate is the volume of the cone changing when the radius is 150 in. and the height is 144 in.?
Answer:
[tex]11304 \frac{in^{3}}{s}[/tex]
Explanation:
r = radius of right circular cone = 150 in
h = height of right circular cone = 144 in
[tex]\frac{dr}{dt}[/tex] = rate at which radius increase = 1.5 in/s
[tex]\frac{dh}{dt}[/tex] = rate at which height decrease = - 2.4 in/s
Volume of the right circular cone is given as
[tex]V = \frac{\pi r^{2}h}{3}[/tex]
Taking derivative both side relative to "t"
[tex]\frac{dV}{dt} = \frac{\pi }{3}\left ( r^{2}\frac{dh}{dt} \right + 2rh\frac{dr}{dt})[/tex]
[tex]\frac{dV}{dt} = \frac{3.14 }{3}\left ( (150)^{2}(- 2.4) + 2(150)(144)(1.5))[/tex]
[tex]\frac{dV}{dt} = 11304 \frac{in^{3}}{s}[/tex]
A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens of the same side as the object. (a) What is the focal length of the lens? Is the lens converging of diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect of inverted?
(a) The focal length of the lens is 6.86 cm, and the lens is converging.
(b) The height of the image is 6.375 mm, and it is inverted.
To solve this problem, we need to use the lens formula and magnification formula, which are given below:
The lens formula is:
[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]
where:
[tex]f[/tex] is the focal length of the lens[tex]d_o[/tex] is the object distance (distance from the object to the lens)[tex]d_i[/tex] is the image distance (distance from the image to the lens)The magnification formula is:
[tex]m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}[/tex]
where:
[tex]m[/tex] is the magnification[tex]h_i[/tex] is the height of the image[tex]h_o[/tex] is the height of the objectNow we can solve the problem step-by-step.
Part (a): Finding the focal length
Given:
Object distance, [tex]d_o = 16.0[/tex] cm
Image distance, [tex]d_i = 12.0[/tex] cm (since the image is on the same side as the object, [tex]d_i[/tex] is positive)
Using the lens formula:
[tex]\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}[/tex]
Substituting the given values:
[tex]\frac{1}{f} = \frac{1}{16.0} + \frac{1}{12.0}[/tex]
Calculating the right-hand side:
[tex]\frac{1}{f} = \frac{3 + 4}{48} = \frac{7}{48}[/tex]
Therefore, the focal length of the lens, [tex]f[/tex], is:
[tex]f = \frac{48}{7} = 6.86[/tex] cm
Since the focal length is positive, the lens is converging.
Part (b): Finding the height and orientation of the image
Given:
Object height, [tex]h_o = 8.50[/tex] mm
Object distance, [tex]d_o = 16.0[/tex] cm
Image distance, [tex]d_i = 12.0[/tex] cm
Using the magnification formula:
[tex]m = -\frac{d_i}{d_o}[/tex]
Substituting the given values:
[tex]m = -\frac{12.0}{16.0} = -0.75[/tex]
Using the magnification formula for height:
[tex]m = \frac{h_i}{h_o}[/tex]
Therefore:
[tex]-0.75 = \frac{h_i}{8.50}[/tex]
Giving:
[tex]h_i = -0.75 \times 8.50 = -6.375[/tex] mm
Since the height of the image, [tex]h_i[/tex], is negative, the image is inverted.
A circular loop 40 cm in diameter is made froma flexible conductor and lies at right angles to a uniform 12-T magnetic field. At time t = 0 the loop starts to expand, its radius increasing at the rate of 5.0 mm/s Find the induced emf in the loop: a) at t 1.0 s and b) at t 10 s.
Answer:
Explanation:
As we know that magnetic flux is given by
[tex]\phi = B.A[/tex]
[tex]\phi = B.\pi r^2[/tex]
now from Faraday's law
[tex]EMF = \frac{d\phi}{dt}[/tex]
[tex]EMF = \frac{d(B. \pi r^2)}{dt}[/tex]
[tex]EMF = 2\pi r B \frac{dr}{dt}[/tex]
now we have
[tex]r = 40/2 = 20 cm[/tex]
B = 12 T
[tex]\frac{dr}{dt} = 5 \times 10^{-3} m/s[/tex]
Part a)
now at t = 1 s
r = 20 + 0.5 = 20.5 cm
[tex]EMF = (2\pi (0.205))(12)(5 \times 10^{-3})[/tex]
[tex]EMF = 0.077 Volts[/tex]
Part b)
now at t = 10 s
r = 20 + 0.5(10) = 25 cm
[tex]EMF = (2\pi (0.25))(12)(5 \times 10^{-3})[/tex]
[tex]EMF = 0.094 Volts[/tex]
A 2.3 kg block is dropped from rest from a height of 4.6 m above the top of the spring. When the block is momentarily at rest, the spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm?
Answer:
7.6 m/s
Explanation:
m = 2.3 kg, h = 4.6 m, x = 25 cm = 0.25 m
Use the conservation of energy
Potential energy of the block = Elastic potential energy of the spring
m g h = 1/2 k x^2
Where, k be the spring constant
2.3 x 9.8 x 4.6 = 0.5 x k x (0.25)^2
k = 3317.88 N/m
Now, let v be the velocity of the block, when the compression is 15 cm.
Again use the conservation of energy
Potential energy of the block = Kinetic energy of block + Elastic potential
energy of the spring
m g h = 1/2 m v^2 + 1/2 k x^2
2.3 x 9.8 x 4.6 = 0.5 x 2.3 x v^2 + 0.5 x 3317.88 x (0.15)^2
103.684 = 1.15 v^2 + 37.33
v = 7.6 m/s
A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. What is the frequency of its fundamental mode of vibration?
B. What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 1.00×10^4 Hz ?
A. 409 Hz
The fundamental frequency of a string is given by:
[tex]f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]
where
L is the length of the wire
T is the tension in the wire
m is the mass of the wire
For the piano wire in this problem,
L = 0.400 m
T = 1070 N
m = 4.00 g = 0.004 kg
So the fundamental frequency is
[tex]f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz[/tex]
B. 24
For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by
[tex]f_n = nf_1[/tex]
where [tex]f_1[/tex] is the fundamental frequency.
Here in this case
[tex]f_1 = 409 Hz[/tex]
A person is capable to hear frequencies up to
[tex]f = 1.00 \cdot 10^4 Hz[/tex]
So the highest harmonics that can be heard by a human can be found as follows:
[tex]f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24[/tex]
The fundamental frequency of the piano wire is approximately 409 Hz, based on the given mass, length, and tension. The highest harmonic that could be heard by a person capable of hearing frequencies up to 10,000 Hz would be the 24th harmonic.
Explanation:To find the fundamental frequency (f1) of the piano wire, we can use the formula for the fundamental frequency of a stretched string, which is f1 = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length (linear mass density).
First, we need to calculate the linear mass density of the piano wire:
Mass (m) = 4.00 g = 0.004 kgLength (L) = 0.400 mLinear mass density (μ) = mass / length = 0.004 kg / 0.400 m = 0.010 kg/mNow, we plug in the values into the formula:
Tension (T) = 1070 NLength (L) = 0.400 mμ = 0.010 kg/mf1 = (1 / (2 * 0.400 m)) * sqrt(1070 N / 0.010 kg/m) = (1 / 0.8 m) * sqrt(107000 N/m) = (1 / 0.8) * 327.16 Hz ≈ 409 Hz.
For part B, we need to determine the highest harmonic that can be heard if the upper limit of human hearing is 10,000 Hz. Since the fundamental frequency is 409 Hz, the harmonics will be integer multiples of this value. The highest harmonic number (n) will be the largest integer such that:
n * f1 ≤ 10,000 Hz
Doing the division gives us n ≤ 10,000 / 409, and we round down to the nearest whole number because we cannot have a fraction of a harmonic. So, the highest harmonic heard would be:
n = floor(10,000/409) = 24 (since 24 * 409 ≈ 9816 Hz which is below the 10,000 Hz threshold).
A 2.04 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.28 kΩ resistor. (a) Calculate the current in the resistor 9.00 µs after the resistor is connected across the terminals of the capacitor.
Answer:
The current in the resistor is 56.44 mA.
Explanation:
Given that,
Capacitor Q= 2.04 nF
Initial charge [tex]q= 4.55\ \mu C[/tex]
Resistor [tex] R= 1.28 k\omega[/tex]
Time [tex]t = 9.00\times10^{-6}[/tex]
We need to calculate the current
Using formula of current
[tex]I=\dfrac{Q}{RC}e^{\dfrac{-t}{RC}}[/tex]
Where, C = capacitor
R = resistor
t = time
Q = charge
Put the value into the formula
[tex]I=\dfrac{4.55\times10^{-6}}{1.28\times10^{3}\times2.04\times10^{-9}}e^{\dfrac{-9.00\times10^{-6}}{1.28\times10^{3}\times2.05\times10^{-9}}}[/tex]
[tex]I=0.05644\ A[/tex]
[tex]I=56.44\ mA[/tex]
Hence, The current in the resistor is 56.44 mA.
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
[tex]s=ut+0.5at^2 \\ [/tex]
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
[tex]h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9 [/tex]
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
[tex]h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\ [/tex]
Solving both equations
[tex]600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\ [/tex]
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
[tex]h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\ [/tex]
Passing of B occurs at 4108.31 height.
An undamped 1.23 kg horizontal spring oscillator has a spring constant of 37.4 N/m. While oscillating, it is found to have a speed of 2.48 m/s as it passes through its equilibrium position. What is its amplitude of oscillation?
Answer:
0.45 m
Explanation:
m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s
velocity is maximum when it passes through the mean position.
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi \sqrt{\frac{1.23}{37.4}}[/tex]
T = 1.139 sec
w = 2 π / T
w = 2 x 3.14 / 1.139
w = 5.51 rad / s
Vmax = w A
Where, A be the amplitude
2.48 = 5.51 A
A = 2.48 / 5.51 = 0.45 m
Final answer:
The amplitude of oscillation for the given spring oscillator is 0.58 m.
Explanation:
The amplitude of oscillation can be calculated using the equation:
A = vmax/ω
where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.
The angular frequency can be calculated using the equation:
ω = sqrt(k/m)
where k is the spring constant and m is the mass of the oscillator.
Substituting the given values into the equations:
ω = sqrt(37.4 N/m / 1.23 kg) = 4.29 rad/s
A = 2.48 m/s / 4.29 rad/s = 0.58 m
Therefore, the amplitude of oscillation is 0.58 m.
A helium nucleus contains two protons and two neutrons. The mass of the helium nucleus is greater than the combined masses of two protons and two neutrons because binding energy has been added. True False
Answer:
False
Explanation:
Actually, the converse is true. The mass number would be lower than the sum of the mass of the individual nucleons combined. According to Einstein’s equation of E=MC², this will be due to a phenomenon called mass defect. This ‘anomaly’ is due to the loss of some energy (now the nuclear binding energy) when the nucleons were brought in together to form the nucleus.
A flashlight bulb operating at a voltage of 14.4 V has a resistance of 11.0 Ω . How many electrons pass through the bulb filament per second (e = 1.6 ´ 10-19 C)? (Give your answer to two significant figures)
For a DC circuit, the following equation relates the voltage, resistance, and current:
V = IR
V is the total voltage supplied, I is the total current, and R is the total resistance.
Given values:
V = 14.4V
R = 11.0Ω
Plug in the values and solve for I:
14.4 = I×11.0
I = 1.309A
Since one electron carries 1.6×10⁻¹⁹C of charge, divide the current by this number.
1.309/(1.6×10⁻¹⁹) = 8.182×10¹⁸
Round this value to 2 significant figures:
8.2×10¹⁸ electrons per second.
A ball is thrown upward from a height of 880 feet above the ground, with an initial velocity of 96 feet per second. From physics it is known that the velocity at time t is v left parenthesis t right parenthesis equals 96 minus 32 t feet per second. a) Find s(t), the function giving the height of the ball at time t. b) How long will the ball take to reach the ground
Answer:
(a): s(t)= hi + Vo * t - g* t²/2
(b): Will take the ball to reach the ground t= 11 seconds.
Explanation:
hi= 880 ft
Vo= 96 ft/s
g= 32 ft/s²
equating to 0 the equation of s(t) and clearing t, we find the time it takes for the ball to fall to the ground.
Answer:
a) [tex]s(t) = 96t - 16t^{2} + 880[/tex]
b) It will take 11 seconds for the ball to reach the ground.
Explanation:
We have an initial height of 880 feet.
And
[tex]v(t) = 96 - 32t[/tex]
a) Find s(t), the function giving the height of the ball at time t
The position, or heigth, is the integrative of the velocity. So
[tex]s(t) = \int {(96 - 32t)} \, dt[/tex]
[tex]s(t) = 96t - 16t^{2} + K[/tex]
In which the constant of integration K is the initial height, so [tex]K = 880[/tex]
So
[tex]s(t) = 96t - 16t^{2} + 880[/tex]
b) How long will the ball take to reach the ground
This is t when [tex]s(t) = 0[/tex]
So
[tex]s(t) = -16t^{2} + 96t + 880[/tex]
This is t = -5 or t = 11.
However, t is the instant of time, so it has to be a positive value.
So it will take 11 seconds for the ball to reach the ground.
A proton is moving at 2.1 x 10 m/s clockwise through a constant and perpendicular magnetic field. The radius of curvature through the field is 0.6 meters. What is the strength of the magnetic field?
Explanation:
For this question the magnetic force provides the force required for the circular motion
Equating
Magnetic force = centripetal force
Bqvsinx = mv^2/r
as magnetic field is perpendicular x=90
Bqvsin90 = mv^2/r
Bq = mv/r
B= mv/rq
Then replace r= 0.6 , v= 2.1ms and mp from formula sheet and you can obtain B which is magnetic field intensity
A disk (radius = 2.00: mm) is attached to a high-speed drill at a dentist's office and is turning at 7.85 times 10^4 rad/s. Determine the tangential speed of a point on the outer edge of this disk.
Answer:
Tangential speed = 157 m/s
Explanation:
Tangential speed = Angular speed x Radius
Angular speed = 7.85 x 10⁴ rad/s
Radius = 2 mm = 2 x 10⁻³ m
Tangential speed = Angular speed x Radius = 7.85 x 10⁴ x 2 x 10⁻³
Tangential speed = 15.7 x 10 = 157 m/s
Tangential speed = 157 m/s
The tangential speed of a point on the outer edge of the disk attached to a high-speed drill, given an angular velocity of 7.85 x 10^4 rad/s and a radius of 2.00 mm, is 157 m/s.
To determine the tangential speed of a point on the outer edge of the disk which is attached to a high-speed drill, we apply the formula that relates tangential speed (v) to the angular velocity ($$omega$) and the radius (r) of the disk: v = r $$omega$. Given the angular velocity of the disk is 7.85 $$x$ 10^4 rad/s and the radius is 2.00 mm (which we need to convert to meters for standard units), the calculation is straightforward.
v = r $$omega$
v = 0.002 m $$x$ 7.85 $$x$ 10^4 rad/s
v = 157 m/s.
Thus, the tangential speed of a point on the outer edge of this disk is 157 m/s.
One characteristic of mass wasting processes is that they a. only operate on steep slopes. b. move materials very slowly. c. operate only during nonfreezing months of the year. d. move materials relatively short distances compared to streams.
Answer:
One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.
An airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 23° south of east. Its direction of motion relative to the earth is 45.0° south of west, while its direction of travel relative to the air is 7° south of west. What is the airplane's speed relative to the air mass in meters per second?
The airplane's speed relative to the air mass is 38.51 m/s. To find the airplane's speed relative to the air mass, you can use the formula sqrt(VAx² + VAy²) by calculating the magnitude of the resultant vector of VAx and VAy.
Explanation:In order to determine the airplane's speed relative to the air mass, a comprehensive analysis involves breaking down velocities into their horizontal and vertical components.
The wind velocity components are established as VWx = magnitude of wind velocity * cos(angle of wind direction) in the 'x' direction and VWy = magnitude of wind velocity * sin(angle of wind direction) in the 'y' direction.
Simultaneously, the airplane's velocity relative to the air mass components are defined as VAx = magnitude of airplane velocity * cos(angle of travel direction relative to air) in the 'x' direction and VAy = magnitude of airplane velocity * sin(angle of travel direction relative to air) in the 'y' direction.
The speed of the airplane relative to the air mass is ascertained by calculating the magnitude of the resultant vector of VAx and VAy using the formula: Speed of airplane relative to air mass = sqrt(VAx² + VAy²).
After substituting the given values in sqrt(VAx² + VAy²): the airplane's speed relative to the air mass is 38.51 m/s.
Therefore, the airplane's speed relative to the air mass is 38.51 m/s.
To move a 51kg cabinet across a floor requires a force of 200N to start it moving, but then only 100N to keep it moving a steady speed. (a) Find the maximum coefficient of static friction between the cabinet and the floor. (b) Find the coefficient of kinetic friction between the cabinet and the floor.
Answer:
a) [tex]\mu_s =0.40[/tex]
b) [tex]\mu_k =0.20[/tex]
Explanation:
Given:
Mass of the cabinet, m = 51 kg
a) Applied force = 200 N
Now the force required to move the from the state of rest (F) = coefficient of static friction ([tex]\mu_s[/tex])× Normal reaction(N)
N = mg
where, g = acceleration due to gravity
⇒N = 51 kg × 9.8 m/s²
⇒N = 499.8 N
thus, 200N = [tex]\mu_s[/tex] × 499.8 N
⇒[tex]\mu_s[/tex] = [tex]\frac{200}{499.8}=0.40[/tex]
b) Applied force = 100 N
since the cabinet is moving, thus the coefficient of kinetic([tex]\mu_k[/tex]) friction will come into action
Now the force required to move the from the state of rest (F) = coefficient of kinetic friction ([tex]\mu_k[/tex])× Normal reaction(N)
N = mg
where, g = acceleration due to gravity
⇒N = 51 kg × 9.8 m/s²
⇒N = 499.8 N
thus, 100N = [tex]\mu_k[/tex] × 499.8 N
⇒[tex]\mu_k[/tex] = [tex]\frac{100}{499.8}=0.20[/tex]
Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water line if it floats in sea water? b) What percentage of the ice floe is below the level of the liquid if it floats in mercury?
Answer:
(a) 92 %
(b) 6.76 %
Explanation:
length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3
density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3
(a) Let v be the volume of ice below water surface.
By the principle of flotation
Buoyant force = weight of ice block
Volume immersed x density of water x g = Total volume of ice block x density
of ice x g
v x 1000 x g = V x 920 x g
v / V = 0.92
% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %
(b) Let v be the volume of ice below the mercury.
By the principle of flotation
Buoyant force = weight of ice block
Volume immersed x density of mercury x g = Total volume of ice block x
density of ice x g
v x 13600 x g = V x 920 x g
v / V = 0.0676
% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %
19. Determine the final state and its temperature when 150.0 kJ of heat are added to 50.0 g of water at 20 ºC. (Specific heat of water = 4.184 J/g ºC; Specific heat of steam = 1.99 J/g • ºC; ∆Hfus (H2O) = 6.01 kJ/mol; ∆Hvap (H2O) = 40.79 kJ/mol). (5 points)
The amount of energy to reach the boiling point is [tex]50*80*4.184 J=16,736J[/tex]. To pass the boiling point, [tex]40.79*\frac{50}{18.02}kJ=113,180J[/tex] are necessary (18.02 is the molar mass of water). This means that [tex]150kJ-113.180kJ-16.736kJ=20,084J[/tex] are left. This allows the steam to heat another [tex]\frac{20,084}{50*1.99}=201.8^{\circ}C[/tex]. Therefore, it ends as steam at temperature [tex]100^{\circ}C+201.8^{\circ}C=301.8^{\circ}C[/tex]
A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10^8 m. How fast does the moon move?
Answer:
v = 895 m/s
Explanation:
Time period is given as 39 Earth Days
[tex]T = 39 days \times 24 hr \times 3600 s[/tex]
[tex]T = 3369600 s[/tex]
now the radius of the orbit is given as
[tex]r = 4.8 \times 10^8 m[/tex]
so the total path length is given as
[tex] L = 2 \pi r[/tex]
[tex]L = 2\pi (4.8 \times 10^8)[/tex]
[tex]L = 3.015 \times 10^9 [/tex]
now the speed will be given as
[tex]v = \frac{L}{T}[/tex]
[tex]v = \frac{3.015 \times 10^9}{3369600} [/tex]
[tex]v = 895 m/s[/tex]
A mass is set oscillating horizontally, with an amplitude of 11.3 cm and a period of 0.702 sec. What is the velocity of the particle when the position is 1/2 the amplitude in the positive x direction and moving toward x=0?
The appropriate solution is "-87.6 cm/s". A complete solution is given below.
According to the question, the given values are:
Amplitude,
a = 11.3 cm
Time,
t = 0.702 sec
→ The velocity (v) will be:
= [tex]\pm w\sqrt{a^2-x^2}[/tex]
or,
= [tex]-\frac{2 \pi}{t} \sqrt{a^2-(\frac{a}{2} )^2}[/tex]
= [tex]-\frac{2 \pi}{t} \frac{\sqrt{3} }{2}a[/tex]
By substituting the given values, we get
= [tex]-\frac{2 \pi}{0.702} \frac{\sqrt{3}\times 11.3 }{2}[/tex]
= [tex]-8.946\times 9.7861[/tex]
= [tex]-87.6 \ cm/s[/tex]
Learn more:
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Final answer:
To find the velocity when the position is half the amplitude in the positive x direction and moving toward x=0, we need to find the time at that position and plug it into the velocity formula.
Explanation:In simple harmonic motion, the velocity is given by the formula v = ωA sin(ωt), where A is the amplitude and ω is the angular frequency. To find the velocity when the position is half the amplitude in the positive x direction and moving toward x=0, we need to find the time at that position and plug it into the velocity formula.
First, we find the time at which the particle is at the given position. Since the period is 0.702 sec, the time at half the amplitude in the positive x direction is 0.351 sec (half the period).
Next, we plug this time into the velocity formula: v = ωA sin(ωt). The angular frequency ω is given by ω = 2π / T, where T is the period. Plugging in the values, we can calculate the velocity.
An engine flywheel initially rotates counterclockwise at 5.85 rotations/s. Then, during 21.3 s, its rotation rate changes to 3.31 rotations/s clockwise. Find the flywheel's average angular acceleration (including its sign) in radians per second squared. Define counterclockwise rotation as positive.
Answer:
- 2.7 rad/s^2
Explanation:
f0 = + 5.85 rotations per second (counter clockwise)
t = 21.3 s
f = - 3.31 rotations per second (clockwise)
w0 = 2 x 3.14 x 5.85 = + 36.738 rad/s
w = - 2 x 3.14 x 3.31 = - 20.79 rad/s
Let α be teh angular acceleration.
α = (w - w0) / t
α = (-20.79 - 36.738) / 21.3
α = - 2.7 rad/s^2
The flywheel's average angular acceleration is -0.43 rad/s².
Average angular accelerationThe average angular acceleration of the flywheel is determined by applying the following kinematic equation as shown below;
α = (ωf - ωi)/t
where;
ωf is the final angular speed = -3.31 rad/sωi is the initial angulra speed = 5.85 rad/st is the time of motion, = 21.3 sα = (-3.31 - 5.85)/21.3
α = -0.43 rad/s²
Thus, the flywheel's average angular acceleration is -0.43 rad/s².
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